# NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4

The statistical branch of Mathematics, which is concerned with the study of the collection, organisation, and interpretation of data, is covered in the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4. Students learn various types of information from a variety of sources. Data refers to the collection of facts and figures that make up this information. The emphasis on information is growing in the world. Every element of people’s lives uses data in some way or another. Statistics is a subject covered in NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4.

The subjects covered in these NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 will help you understand the concepts of data collection and the distinction between primary and secondary data.The several types of graphical data representation that will be taught to the students, along with measurements of central tendency including mean, median, and mode. The NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 are available in PDF format below. Students can also find some of these in the exercises that follow.

• NCERT Solutions Class 9 Maths Chapter 14 Ex 14.1
• NCERT Solutions Class 9 Maths Chapter 14 Ex 14.2
• NCERT Solutions Class 9 Maths Chapter 14 Ex 14.3
• NCERT Solutions Class 9 Maths Chapter 14 Ex 14.4

Questions in the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 will teach students about primary and secondary data, how to organise ungrouped data, and how to determine the mean, median, and mode.

The NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 provide an excellent combination of questions that let students explore a variety of ideas. They will gain a better understanding of the principles if they practise these NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 on a regular basis.These questions are extremely important, since the expert team at NCERT thoroughly researched these NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 before developing them. The NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 are thoroughly evaluated below.

NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 explains the concept of data, the methods for gathering it, the distinction between primary and secondary data, the graphical depiction of data using bar graphs, histograms, and frequency polygons, as well as measures of central tendency like mean, median, and mode, which are all covered in the NCERT Solutions for Class 9 Chapter 14 Exercise 14.4.

There are 26 questions in the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 and most of them are simple formula questions. About two of the theoretical short answer questions are really elementary, 20 are relatively simple when the proper formulas are used and graphs are shown, and the final four can be a little time-consuming.

For mastering Statistics, a variety of key ideas and formulas are included in the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4, similar to how mean, median, and mode are calculated. While “mode” is the most typical value, the term “mean” refers to the average value. Below are a few key equations from the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4:

• Class-mark = (Upper limit + Lower limit)/ 2
• Mean = Sum of the observations / Total number of observations

The Extramarks website and mobile applications both offer NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4. Mathematical textbook solutions are included in the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4.

## NCERT Solutions for Class 9 Maths Chapter 14 Statistics (Ex 14.4) Exercise 14.4

On the Extramarks website and mobile application, students can find the thorough NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4. Students from all around India who are preparing for their Class 9 Mathematics examinations using NCERT textbooks can refer to these NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 offered by Extramarks. Exercises 14.1, 14.2, 14.3, and 14.4 are covered in the Chapter 14 NCERT Solutions for Class 9 Mathematics. Extramarks’ top mathematics experts created these NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4.

Students will be capable of comprehending the concepts and complete their projects and homework on time with the aid of these NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4. Additionally, it will aid in an efficient chapter rewrite. With the use of the link provided on the Extramarks website and mobile application, students can easily access the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 PDF solutions and study offline.

Students in Class 9 must maintain their understanding of their ideas. As a result, using the NCERT Solutions for Class 9 Chapter 14 Exercise 14.4 might be quite beneficial. The students will find it simpler to study for the test. Students are guaranteed to be familiar with the question structure and format by using the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 by Extramarks.

Extramarks offers a variety of questions that will benefit students over time. It is advised that students extensively practise these NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 available on the Extramarks website and mobile application.

### Access NCERT Solutions for Class 9 Maths Chapter 14 – Statistics

Students can conveniently avail the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 on the Extramarks website and mobile application. The NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 are written in an easy to understand language and have detailed explanations of each part of  Class 9 Maths Chapter 14 Exercise 14.4.

Here are a few points students need to note before practising the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4:

• Data simply refers to information. The definition given in the dictionary is “given facts.”
• The frequency of an observation is the number of times the observation appears in the given data.
• A bar graph is a visual display of numerical data that consists of a number of bars (rectangles) that are all the same width and are arranged horizontally or vertically with equal gaps  between them.
• A frequency polygon for a specific frequency distribution is another way to express frequency distribution graphically.
• The observation that occurs most frequently is the mode.

### NCERT Solution Class 9 Maths of Chapter 14 All Exercises

One of the finest ways to improve one’s abilities and knowledge is to use the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4. The NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 includes all the necessary study material that will enable students to perform well on the CBSE exams. Statistics is a crucial chapter and is covered in NCERT Solutions Class 9 Maths Chapter 14 Exercise 14.4 as part of the Class 9 Mathematics CBSE Syllabus 2022–23.

The NCERT Solutions for Class 9  Maths chapter 14 Exercise 14.4 provides answers to the questions in the  NCERT textbook. The NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 are precise and come from subject-matter experts. For the students’ benefit, each question is broken down into steps. The best resource for students preparing for board exams is the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4. The solutions should be read by students after finishing each chapter. They might analyse their weaker areas and as a result, and work towards strengthening them.

The Extramarks NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 are organised in an incredibly effective way. The responses allow students the chance to assess the conventional or evident manner of answering a question, reflecting the order of chapters and subjects in the textbooks. The NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4  are ingenious since they get to the conclusion in the best way and avoid confusing students. The language is simple, reducing the students’ time and preparation effort.

Apart from the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 provided by Extramarks, here is a list of other exercises that Extramarks provides NCERT Solutions for:

The Collection of Data is covered in the first section. An exercise is used to explain this specific area. There are only two questions in Exercise 14.1.

The next topic covered in the same chapter is Presentation of Data under Exercise 14.2. Students must keep these things in mind before solving the NCERT Solutions for this exercise:

• Raw data, range, frequency, an ungrouped frequency distribution table, and tally marks are other terminology that students must understand.
• Classes or class intervals are the units of analysis described in this chapter. Class size or class width refers to the dimensions of a class.
• The lowest number in each of these classes is referred to as the lower class limit, and the highest number is referred to as the upper class limit; for example, in the class of 20 to 29, 20 is the “lower class limit” and 29 is the “upper class limit.”

Students will find examples with answers and practise questions in this topic to help them better understand the concept.

The following is an explanation of graphical data representation.This comprises:

• Bar graphs
• Histograms with varied and uniform widths. Similar to a bar graph, this type of depiction is utilised for continuous class intervals.
• Number of polygons.
• Additionally, frequency polygons may be created independently without histograms.
• Class-marks are the midpoints of the class intervals.
• Class-mark is equal to (Upper limit plus Lower limit) / 2.

Several open-ended questions are included in Exercise 14.3 for the purpose of student evaluation. The final subject covered in this chapter is the central tendency measure. For ungrouped data, the three measures of the central tendency are:

• The mean is calculated by adding up all the observational data and dividing that total by the number of observations.
• The value of the middle observation is known as the median (s).
• Mode: The mode is the observation that occurs most frequently.

### NCERT Solutions for Class 9 Chapter 14 Statistics Exercise 14.4

Students will converse about brand-new subjects and ideas in this part. Since exercise 14.3 was the previous exercise, it is known that statistics covers a wide range of crucial concepts in just one chapter of Mathematics, including frequency, range, and many graph types. Examining and analysing additional material is a continuation of the chapter’s exercise 14.2, which featured numerical questions based on frequency and range. It is found in the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4. The ideas of mean, median, and mode will be studied by the students in the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4. Students will be examining all of its uses consecutively, starting with the formula and ending with how it will be used.

In addition to theory, the NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 also offers numerical exercises and takes into consideration new concepts such as questions based on observations and maximum and lowest projected values that are crucial to it. Only real-world examples that are significant will be covered in NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.4 are included in this NCERT syllabus Chapter.

### NCERT Solutions for Class 9

Sophisticated concepts at higher levels can be understood thanks to the notions covered in Class 9. As a result, the NCERT curriculum for Science, Math, and Social Science is more complicated than in prior years and demands greater concentration. The subjects covered in NCERT Class 9 are crucial for competitive exams.

To enable students to achieve the highest possible grades in the exams, Extramarks has provided NCERT Solutions for Class 9 for all subjects. For the exercises found in the back of the Class 9 NCERT textbooks, subject specialists at Extramarks have developed these solutions. They were created in a thorough, step-by-step manner in accordance with the most recent recommendations.

### CBSE Study Materials for Class 9

Top academic professionals at Extramarks have solved the NCERT Class 9 Solutions for the major subjects, which are available on the Extramarks website and mobile application. These chapter-by-chapter NCERT Solutions rigorously follow the most recent regulations and grading scales to aid students in performing well on exams. Students can use these solutions if they run into problems when attempting to answer in-text questions.

Students will have a better knowledge of the associated ideas thanks to these step-by-step solutions, which were prepared by qualified specialists. They will assist pupils in removing their uncertainties and enhancing their problem-solving skills.

Since CBSE Class 9 can be thought of as the foundation for higher classes, it is crucial that students understand the material completely. The lesson plans are created so that examinees can learn and grow in their sense of individuality, which inevitably affects their future. The board and NCERT jointly released the rules for CBSE Class 9. Thus, the NCERT curriculum is used at schools that are linked with the CBSE board. Due to the fact that these topics will be covered in higher standards, it is crucial that students learn these subjects like Science and Mathematics correctly by comprehending each concept and topic. Understanding the curriculum is crucial for effective preparation, as is working through a variety of sample problems.

### CBSE Study Materials

One of the most well-known and eminent educational boards in the nation is the Central Board of Secondary Education, or CBSE for short. It is a part of the Union Government of India. Numerous public and private schools that are part of the CBSE board use the NCERT curriculum.

Q.1 The following number of goals were scored by a team in a series of 10 matches:

2, 3, 4, 5, 0, 1, 3, 3, 4, 3

Find the mean, median and mode of these scores.

Ans

$\begin{array}{l}\text{Mean=}\frac{\mathrm{Sum}\mathrm{of}\mathrm{data}}{\mathrm{Number}\mathrm{of}\mathrm{data}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2+3+4+5+0+1+3+3+4+3}{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{28}{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2.8\\ \text{Thus},\text{the mean of the score is 2}\text{.8}\text{.}\\ \text{Ascending order of goals:}\\ 0,\text{1, 2, 3, 3, 3, 3, 4, 4, 5}\\ \text{Number of observations}\left(n\right)=10\text{\hspace{0.17em}}\left(even\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}Median=\frac{{\left(\frac{n}{2}\right)}^{th}term+{\left(\frac{n}{2}+1\right)}^{th}term}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{{\left(\frac{10}{2}\right)}^{th}term+{\left(\frac{10}{2}+1\right)}^{th}term}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{{\left(5\right)}^{th}term+{\left(6\right)}^{th}term}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3+3}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3\\ \text{Thus, median of the goals is 3}\text{.}\\ \text{Mode:}\\ \text{Since},\text{the frequency of 3 is more}\text{.}\\ \text{So, mode of goals scored by the team}=\text{3}\text{.}\end{array}$

Q.2 In a mathematics test given to 15 students, the following marks (out of 100) are recorded:
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.

Ans

$\begin{array}{l}\text{Mean of the marks obtained in test}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\begin{array}{l}\text{41}+\text{39}+\text{48}+\text{52}+\text{46}+\text{62}+\text{54}+\text{4}0+\text{96}+\text{52}+\text{98}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{4}0+\text{42}+\text{52}+\text{6}0\end{array}}{15}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{822}{15}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=54.8\\ \text{Ascending order of marks:}\\ \text{39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98}\\ \mathrm{n}=15\left(\mathrm{odd}\right)\\ \text{∴Median}={\left(\frac{\mathrm{n}+1}{2}\right)}^{\mathrm{th}}\mathrm{term}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(\frac{15+1}{2}\right)}^{\mathrm{th}}\text{\hspace{0.17em}}\mathrm{term}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}={8}^{\mathrm{th}}\text{\hspace{0.17em}}\mathrm{term}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=52\\ \text{Here},\text{the frequency of 52 is heighest i.e., 3.}\\ \text{So, the mode is 52.}\end{array}$

Q.3 The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95

Ans

$\begin{array}{l}\mathrm{The}\text{given data are:}\\ \text{29},\text{32},\text{48},\text{5}0,\text{}\mathrm{x},\text{}\mathrm{x}+\text{2},\text{72},\text{78},\text{84},\text{95}\\ \text{Number of data}=10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\mathrm{Median}=63\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{Median}=\frac{{\left(\frac{\mathrm{n}}{2}\right)}^{\mathrm{th}}\mathrm{term}+{\left(\frac{\mathrm{n}}{2}+1\right)}^{\mathrm{th}}\mathrm{term}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\left(\frac{10}{2}\right)}^{\mathrm{th}}\mathrm{term}+{\left(\frac{10}{2}+1\right)}^{\mathrm{th}}\mathrm{term}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}63=\frac{{5}^{\mathrm{th}}\mathrm{term}+{6}^{\mathrm{th}}\mathrm{term}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}63=\frac{{5}^{\mathrm{th}}\mathrm{term}+{6}^{\mathrm{th}}\mathrm{term}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}63=\frac{\mathrm{x}+\mathrm{x}+2}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}63=\frac{2\mathrm{x}+2}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}63=\mathrm{x}+1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=63-1=62\end{array}$

Q.4 Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.

Ans

$\begin{array}{l}Ascending\text{order of data is:}\\ \text{14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28}\\ \text{Here 14 is repeated more time than any other data}\text{.}\\ \text{So, mode of data is 14}\text{.}\end{array}$

Q.5 Find the mean salary of 60 workers of a factory from the following table:

 Salary (in Rs) Number of workers 3000 4000 5000 6000 7000 8000 9000 10000 16 12 10 8 6 4 3 1 Total 60

Ans

 Salary(x) Number of workers(f) fx 3000 4000 5000 6000 7000 8000 9000 10000 16 12 10 8 6 4 3 1 48000 48000 50000 48000 42000 32000 27000 10000 Total 60 305000

$\begin{array}{l}\mathrm{Mean}=\frac{\sum \mathrm{fx}}{\sum \mathrm{f}}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{305000}{60}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5083.33\end{array}$

Q.6 Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.

Ans

(i) Here is an example showing that mean is an appropriate measure of central tendency. The heights of five friends: 110.8 cm, 110.9 cm, 111.1 cm, 111.2 cm and 111.4 cm.

$\begin{array}{l}\text{Then},\text{Mean height}=\frac{\begin{array}{l}\text{11}0.\text{8 cm}+\text{11}0.\text{9 cm}+\text{111}.\text{1 cm}+\text{111}.\text{2 cm}\\ \text{}+\text{111}.\text{4 cm}\end{array}}{5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{555.4}{5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}=111.08\text{\hspace{0.17em}}cm\end{array}$

We see that mean height is approximately 111 cm which is closer to each given data.
So, we can say that mean is an appropriate measure of central tendency.
(ii) Example of median:

$\begin{array}{l}\mathrm{The}\text{temperatures in °C of 6 different cities are given below:}\\ \text{26°,\hspace{0.17em}\hspace{0.17em}28°,\hspace{0.17em}\hspace{0.17em}29°,\hspace{0.17em}\hspace{0.17em}32°,\hspace{0.17em}\hspace{0.17em}35°,\hspace{0.17em}\hspace{0.17em}41°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Mean}=\frac{\text{26°}+\text{\hspace{0.17em}\hspace{0.17em}28°}+\text{\hspace{0.17em}\hspace{0.17em}29°}+\text{\hspace{0.17em}\hspace{0.17em}32°}+\text{\hspace{0.17em}35°}+\text{\hspace{0.17em}\hspace{0.17em}41°}}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{191\mathrm{°}}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=31.8\mathrm{°}\\ \text{Here, n}=6\left(\mathrm{even}\right)\\ \mathrm{So},\text{median}=\frac{{\left(\frac{\mathrm{n}}{2}\right)}^{\mathrm{th}}\text{term}+{\left(\frac{\mathrm{n}}{2}+1\right)}^{\mathrm{th}}\text{term}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{{\left(\frac{6}{2}\right)}^{\mathrm{th}}\text{term}+{\left(\frac{6}{2}+1\right)}^{\mathrm{th}}\text{term}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{3}^{\mathrm{rd}}\text{term}+{4}^{\mathrm{th}}\text{term}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{29°}+\text{\hspace{0.17em}\hspace{0.17em}32°}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{61°}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=30.5\mathrm{°}\\ \mathrm{We}\text{see that median is better representation than mean.}\\ \text{Because, Meadian does not affect the heighest temperature.}\end{array}$